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(18) A body starts its motion in a constant direction with a velocity 15 cm/sec and uniform acceleration 4 cm/sec in the same direction as its velocity.
Find:
(1) The distance covered in the sixth second only.
(2nd) The distance covered in the seventh and the eighth seconds only. ​

Respuesta :

sqdancefan

Answer:

  1. 37 cm
  2. 86 cm

Step-by-step explanation:

Given a body starts with a velocity of 15 cm/s and an acceleration of 4 cm/s², you want (1) the distance covered in the 6th second, and (2) the distance covered in the 7th and 8th seconds.

Distance

The speed is the sum of the initial velocity and the increase due to acceleration:

  v = 15 +4t . . . . . cm/s after t seconds

The distance is the integral of the velocity.

  [tex]\displaystyle s =\int_{t_0}^{t_1}{(15 +4t)}\,dt=15(t_1-t_0)+2(t_1^2-t_0^2)\\\\s=(t_1-t_0)(2(t_1+t_0)+15)[/tex]

1. 6th second

The sixth second is the one ending at t=6, so the distance is ...

  s = (6 -5)(2(6 +5) +15) = 1(2(11) +15) = 37

The distance covered in the 6th second is 37 cm.

2. 7th and 8th seconds

This is the time period between t=6 and t=8, so the distance is ...

  s = (8 -6)(2(8 +6) +15) = 2(2(14) +15) = 86

The distance covered in the 7th and 8th seconds is 86 cm.

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